Problem: The lifespans of porcupines in a particular zoo are normally distributed. The average porcupine lives $21$ years; the standard deviation is $1.6$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a porcupine living less than $16.2$ years.
$21$ $19.4$ $22.6$ $17.8$ $24.2$ $16.2$ $25.8$ $99.7\%$ $0.15\%$ $0.15\%$ We know the lifespans are normally distributed with an average lifespan of $21$ years. We know the standard deviation is $1.6$ years, so one standard deviation below the mean is $19.4$ years and one standard deviation above the mean is $22.6$ years. Two standard deviations below the mean is $17.8$ years and two standard deviations above the mean is $24.2$ years. Three standard deviations below the mean is $16.2$ years and three standard deviations above the mean is $25.8$ years. We are interested in the probability of a porcupine living less than $16.2$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the porcupines will have lifespans within 3 standard deviations of the average lifespan. The remaining $0.3\%$ of the porcupines will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({0.15\%})$ will live less than $16.2$ years and the other half $({0.15\%})$ will live longer than $25.8$ years. The probability of a particular porcupine living less than $16.2$ years is ${0.15\%}$.